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Question

The equation of the circle passing through points of intersection of the circle x2+y22x4y+4=0 and the line x+2y=4 and touches the line x+2y=0, is

A
x2+y2+x+2y=0
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B
x2+y2+x2y=0
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C
x2+y2x2y=0
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D
x2+y2x+2y=0
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Solution

The correct option is C x2+y2x2y=0
Equation of any circle passing through points of intersection of the given circle and the line is
(x2+y22x4y+4)+λ(x+2y4)=0
x2+y2+(λ2)x+(2λ4)y+4(1λ)=0(1)

It will touch the line x+2y=0 if solution of equation (1) and x+2y=0 is unique.

Hence the roots of the equation
(2y)2+y2+(λ2)(2y)+(2λ4)y+4(1λ)=0
5y2+4(1λ)=0 must be equal.
04×5×4(1λ)=0
λ=1
From (1), the required circle is x2+y2x2y=0

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